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            <div class="quiz-header">
              <h1>Java中的无锁编程 - 交互式测验</h1>
              <p>检验你对文档内容的理解程度</p>
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  {
    "question": "根据文档，Java无锁编程主要依赖于哪种核心技术来实现线程安全的数据更新，以避免传统锁带来的开销？",
    "options": [
      "`ReentrantLock`提供的可重入锁。",
      "CAS (Compare-And-Swap) 操作，一种硬件支持的原子操作。",
      "线程池的任务调度机制。",
      "`synchronized`关键字提供的互斥锁。"
    ],
    "correctAnswerIndex": 1,
    "explanation": "文档明确指出，无锁编程主要通过CAS（比较并交换）操作实现。CAS是一种硬件支持的原子操作，它可以在不加锁的情况下检查并更新数据状态，从而避免线程阻塞和锁竞争。Java中的原子类如`AtomicInteger`就是基于此技术实现的。",
    "source": "文档中「无锁编程的核心技术：CAS」部分",
    "hint": "文档提到，这种技术是硬件支持的原子操作，并且是Java中原子类的实现基础。"
  },
  {
    "question": "在无锁编程中，\"ABA问题\"指的是什么？",
    "options": [
      "两个线程A和B互相等待对方释放资源而导致的死锁。",
      "一个操作因为高并发冲突而反复失败，导致CPU资源被大量消耗。",
      "线程在尝试获取资源时，因为优先级反转而导致长时间等待。",
      "一个值从A变为B，再变回A，导致CAS操作错误地认为该值从未改变。"
    ],
    "correctAnswerIndex": 3,
    "explanation": "ABA问题是指在CAS操作期间，一个内存位置的值最初是A，被其他线程修改为B，然后又被改回A。当执行CAS的线程检查时，发现值仍然是A，便错误地认为没有发生过变化，从而可能导致数据不一致。",
    "source": "文档中「无锁编程的挑战」部分",
    "hint": "这个问题描述了一个变量的值发生了“看似未变”的变化过程，从而可能欺骗CAS操作。"
  },
  {
    "question": "文档中提到了哪种类来专门解决CAS操作中的\"ABA问题\"？",
    "options": [
      "`AtomicStampedReference`，通过引入版本号或标记来检查数据是否被篡改。",
      "`LongAdder`，通过分散计数来减少竞争。",
      "`AtomicInteger`，通过原子自增来避免冲突。",
      "`ConcurrentHashMap`，通过分段锁来隔离数据。"
    ],
    "correctAnswerIndex": 0,
    "explanation": "文档明确指出，解决ABA问题的常用方法是使用“版本号”或“标记”机制。`AtomicStampedReference`就是通过附加一个版本号（stamp）来实现的，CAS操作时会同时检查值和版本号，从而确保数据的一致性。",
    "source": "文档中「4. 版本号和标记机制（解决ABA问题）」部分",
    "hint": "思考一下哪种机制除了比较值本身，还能额外比较值的“版本”。"
  },
  {
    "question": "为什么在高并发计数场景下，`LongAdder`的性能通常优于`AtomicLong`？",
    "options": [
      "`LongAdder`的所有操作都是由操作系统内核直接执行，绕过了JVM。",
      "`LongAdder`使用了更复杂的锁机制，保证了绝对的线程安全。",
      "`AtomicLong`在每次操作后都需要进行一次完整的垃圾回收。",
      "`LongAdder`将计数操作分散到多个独立的计数段上，极大地减少了CAS冲突。"
    ],
    "correctAnswerIndex": 3,
    "explanation": "`LongAdder`的核心优势在于其“分段计数”的设计。它将单一的计数值分散到多个内部变量上，不同线程可以更新不同的段，从而在高并发环境下大大减少了对同一个变量进行CAS操作的竞争，提升了吞吐量。",
    "source": "文档中「6. LongAdder 和 DoubleAdder（高并发计数器）」部分",
    "hint": "文档中提到了一个关键词：“分段计数”，这是理解其高性能的关键。"
  },
  {
    "question": "在无锁编程中，当多个线程在高冲突场景下反复尝试CAS操作失败时，会产生什么问题？文档中推荐了哪种策略来缓解此问题？",
    "options": [
      "产生“忙等”问题；推荐采用指数级回退策略。",
      "产生内存泄漏问题；推荐手动调用垃圾回收。",
      "产生死锁问题；推荐使用`synchronized`来解决。",
      "产生ABA问题；推荐使用`AtomicStampedReference`来解决。"
    ],
    "correctAnswerIndex": 0,
    "explanation": "高冲突下线程不断重试CAS操作会浪费大量CPU资源，这种现象被称为“忙等”（Busy-Waiting）。文档推荐使用“指数级回退策略”，即让线程在每次重试失败后等待一个逐渐增长的时间，以减少重试频率和CPU消耗。",
    "source": "文档中「5. 指数级回退（Exponential Backoff）策略（解决忙等问题）」部分",
    "hint": "这个问题关注的是高冲突下不断重试操作所带来的CPU资源浪费，以及如何让线程“聪明地”等待。"
  },
  {
    "question": "在Java无锁编程中，`volatile`关键字扮演了什么重要角色？",
    "options": [
      "它能自动将CAS操作失败的线程置于休眠状态，以节省CPU资源。",
      "它的主要作用是标记变量，以便垃圾回收器优先回收。",
      "它提供了互斥锁功能，确保同一时间只有一个线程能修改变量。",
      "它能确保变量的内存可见性，并防止指令重排，从而保证CAS操作的正确性。"
    ],
    "correctAnswerIndex": 3,
    "explanation": "`volatile`关键字在并发编程中有两大作用：一是保证变量修改后对其他线程的可见性；二是禁止指令重排序优化。这两点对于依赖最新值进行比较和交换的CAS操作至关重要，是保证无锁算法正确性的基础。",
    "source": "文档中「6. JMM（Java Memory Model）和`volatile`的结合」部分",
    "hint": "想想`volatile`在Java内存模型（JMM）中的两个核心保证：可见性和有序性。"
  },
  {
    "question": "根据文档，无锁编程的一个主要缺点或挑战是什么？",
    "options": [
      "实现逻辑通常比传统的加锁方式更简单直观。",
      "无法在多核CPU上运行，可扩展性差。",
      "实现复杂度高，且在写操作频繁的场景下可能因高重试成本导致性能下降。",
      "必然会导致死锁问题，而有锁编程可以轻易避免。"
    ],
    "correctAnswerIndex": 2,
    "explanation": "文档在“缺点”部分明确指出，无锁编程的实现复杂度高，对开发人员要求高。此外，在写操作频繁的场景下，频繁的CAS重试会消耗大量CPU资源（即高重试成本），反而可能导致性能不如传统锁机制。",
    "source": "文档中「无锁编程的优缺点」部分",
    "hint": "思考一下在什么情况下，无锁编程的“乐观”尝试会变成一种负担。"
  }
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                '</div>' +
                '</div>' +
                '<div class="report-actions">' +
                '<button id="review-btn">回顾测验</button>' +
                '<button class="primary" onclick="location.reload()">再试一次</button>' +
                '</div>' +
                '</div>';

            let reportEl = document.querySelector('.final-score-container');
            if (reportEl) {
                reportEl.remove();
            }
            quizContentEl.insertAdjacentHTML('beforeend', finalScoreHtml);
            document.getElementById('review-btn').addEventListener('click', reviewQuiz);
        }

        function reviewQuiz() {
            const reportEl = document.querySelector('.final-score-container');
            if (reportEl) {
                reportEl.remove();
            }

            document.querySelector('.quiz-body').style.display = 'block';
            document.getElementById('quiz-footer').style.display = 'flex';
            
            currentQuestionIndex = 0;
            loadQuestion();
            updateNavigationButtons(); // Ensure nav buttons are correctly displayed
        }

        // Event Delegation for options
        optionsListEl.addEventListener('click', (event) => {
            if (event.target && event.target.matches('li.option-item')) {
                // Check if the question has already been answered
                if (userAnswers[currentQuestionIndex] === null) {
                    const selectedIndex = parseInt(event.target.dataset.index, 10);
                    selectOption(selectedIndex);
                }
            }
        });

        prevBtn.addEventListener('click', goToPreviousQuestion);
        nextBtn.addEventListener('click', goToNextQuestion);
        hintBtn.addEventListener('click', showHint);
        
        // 初始加载
        loadQuestion();
      </script>
    </body>
    </html>
  